Free $S^1$-action on compact homogeneous spaces Let $M = G/K$ be a compact homogeneous space, i.e. $G$ a connected compact Lie group, and $K$ a closed subgroup inside $G$ that contains no nontrivial normal subgroup of $G$.

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*If $r(G) > r(K)$ (here, $r$ means the rank of maximal torus), is it true that $G/K$ admits free $S^1$-action? If so, why? If not, what is a counterexample?

*What if we further assume $K$ is connected?
(Note: In “Homogeneous Spaces with Non-Vanishing Euler Characteristics” by HC Wang (1949), it was proven that $\chi(G/K) \ne 0 \Leftrightarrow r(G) = r(K)$.)

*If 1. is unknown, then, restricting to case when $G$ semisimple, is 1. known?

Any input would be really helpful. Thanks a lot!
 A: Here is a counterexample where $K$ is connected.
Let $W = SU_3/SO_3$ be the Wu manifold. This 5-dimensional manifold is simply-connected and has $H_2(W;\Bbb Z) = \Bbb Z/2$, and in particular, $\pi_2(W) = \Bbb Z/2$ by the Hurewicz theorem. Note that $r(SU_3) = 2$ and $r(SO_3) = 1$.
Now I claim that there is no free circle action on $W$. For if $M$ is any manifold with free circle action and quotient $X$, the projection defines a fiber-bundle $$S^1 \to M \to X.$$

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*If $M$ is simply-connected, the long exact sequence of homotopy groups implies $X$ is, too.

*If $M$ is further a simply-connected 5-manifold, so that $X$ is now a simply-connected 4-manifold, observe that by Poincare duality $H_2(X;\Bbb Z)$ is free abelian. The long exact sequence of homotopy groups gives an injection of $H_2(W;\Bbb Z) \cong \pi_2(W)$ into $\pi_2(X) \cong H_2(X;\Bbb Z)$. Thus if $M$ is a simply-connected 5-manifold which supports a free circle action, $\pi_2(M)$ must be free abelian.

Because the Wu manifold has $\pi_2(W) = \Bbb Z/2$, it does not support a free circle action. This answers your question in the negative, and unfortunately I do not see a way to salvage it by adding more conditions.
